数学建模小作业

1.某厂生产甲乙两种口味的饮料,每百箱甲饮料需用原料6千克,工人10名,可获利10万元;每百箱乙饮料需用原料5千克,工人20名,可获利9万元.今工厂共有原料60千克,工人150名,又由于其他条件所限甲饮料产量不超过8百箱.问如何安排生产计划,即两种饮料各生产多少使获利最大. model: sets: si/1..3/:B; SJ/1..2/:c,x; sij(si,sj):A; end sets data: B=60 150 8; A=6 5 10 20 1 0; C=10 9; end data

max=@sum(sj:C*X);

@for(si(i):@sum(sj(j):A(i,j)*x(j))

Global optimal solution found.

Objective value: 98.00000 Objective bound: 98.00000 Infeasibilities: 0.000000 Extended solver steps: 0 Total solver iterations: 4

Variable Value Reduced Cost B( 1) 60.00000 0.000000 B( 2) 150.0000 0.000000 B( 3) 8.000000 0.000000 C( 1) 10.00000 0.000000 C( 2) 9.000000 0.000000 X( 1) 8.000000 -10.00000 X( 2) 2.000000 -9.000000 A( 1, 1) 6.000000 0.000000 A( 1, 2) 5.000000 0.000000 A( 2, 1) 10.00000 0.000000 A( 2, 2) 20.00000 0.000000

A( 3, 1) 1.000000 0.000000 A( 3, 2) 0.000000 0.000000

Row Slack or Surplus Dual Price 1 98.00000 1.000000 2 2.000000 0.000000 3 30.00000 0.000000 4 0.000000 0.000000

2.某厂向用户提供发动机，合同规定，第一、二、三季度末分别交货40台、60台、80台．每季度的生产费用为 （元），其中x是该季生产的台数．若交货后有剩余，可用于下季度交货，但需支付存储费，每台每季度c元．已知工厂每季度最大生产能力为100台，第一季度开始时无存货，设a=50、b=0.2、c=4，问工厂应如何安排生产计划，才能既满足合同又使总费用最低．

初次：但有误 model: sets:

M/1..3/:x,c,d; 少endsets data: a=50 ; b=0.2 ; c=4 4 4; d=40 60 80; e=100 ; end data

min=@sum(M:a*x)+@sum(M:b*x*x)+@sum(M(j)|i#le#2:@sum(M(i)|i#le#j:x-d)*c(j+1));

@for(M(j)|j#le#2:x>d); @sum(M:x)=@sum(M:d); @for(M:x

M/1..3/:x,c,d; endsets data: a=50; b=0.2;

c=4 4 4; d=40 60 80; e=100 ; enddata

min=@sum(M:a*x)+@sum(M:b*x*x)+@sum(M(j)|j#le#2:@sum(M(i)|i#le#j:x-d)*c(j+1));

@for(M(j)|j#le#2:x>d); @sum(M:x)=@sum(M:d); @for(M:x

Local optimal solution found.

Objective value: 11280.00 Infeasibilities: 0.5684342E-13 Extended solver steps: 2 Total solver iterations: 14

Variable Value Reduced Cost A 50.00000 0.000000 B 0.2000000 0.000000 E 100.0000 0.000000 X( 1) 50.00000 0.000000 X( 2) 60.00000 0.000000 X( 3) 70.00000 0.000000 C( 1) 4.000000 0.000000 C( 2) 4.000000 0.000000 C( 3) 4.000000 0.000000 D( 1) 40.00000 0.000000 D( 2) 60.00000 0.000000 D( 3) 80.00000 0.000000

Row Slack or Surplus Dual Price 1 11280.00 -1.000000 2 10.00000 0.000000 3 0.000000 0.000000 4 0.000000 -78.00000 5 50.00000 0.000000 6 40.00000 0.000000 7 30.00000 0.000000

3.某商店有五位工作人员:经理1人,主任1人,售货员3人.有关情况见下表.设广告费对销售额的贡献为其投入的15倍,各工作人员的收入相当于其完成销售额的5.5%.问如何安排才能达到以下的目标:P1保证全体人员正常工作时间;P2 至少完成销售额70000元;P3主任的月收入不少于1200元,售货员A和B的月收入不少于600元和400元;P4 全体人员加班时间不超过规定; P5广告费不超过3000元,力争销售额增加10000元,前者的重要性为后者的两倍.

经理 主任 每小时对销售额的贡献每月总工时 每月加班限量(工(元) 时) 144 96 200 24 200 24 172 52 160 32 100 32 售货员 54 A 售货员B 30 售货员C 9

model: sets:

sj/1..16/:B; sk/1..38/:y; sjk(sj,sk):A;

end sets data:

B=200 200 172 160 100 70000 1200 600 400 224 224 224 192 132 3000 10000; A=1 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 1 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 1 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 1 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0

0 0 0 0 1 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0

144 96 54 30 9 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0

0 5.28 0 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0

0 0 2.97 0 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0

0 0 0 1.65 0 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0

1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0

0 0 0 0

0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0

0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0

0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0

0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0

0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0

0 0 0 0 0 15 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1;

end data

min=@sum(sk(k)|(k#ge#23)#and#(k#le#27):y(k)*0.5)+y(28)*0.2+@sum(sk(k)|(k#ge#28)#and#(k#le#30):y(k)*0.15)+@sum(sk(k)|(k#ge#16)#and#(k#le#20):y(k)*0.09)+(y(21)*2+y(38)*0.06);

@for(sj(j):@sum(sk(k):A(j,k)*y(k))=B(j));

@for(sk(k)|k#le#5:@gin(y(k))); End

Solve：

Global optimal solution found.

Objective value: 0.3600000 Objective bound: 0.3600000 Infeasibilities: 0.000000 Extended solver steps: 0 Total solver iterations: 11

Variable Value Reduced Cost B( 1) 200.0000 0.000000 B( 2) 200.0000 0.000000 B( 3) 172.0000 0.000000 B( 4) 160.0000 0.000000 B( 5) 100.0000 0.000000 B( 6) 70000.00 0.000000 B( 7) 1200.000 0.000000